MG6863 - ENGINEERING ECONOMICS

PART – A (2 Marks)

UNIT – I: INTRODUCTION TO ECONOMICS

1. What is elasticity of demand? Elasticity of demand may be defined as „the degree of responsiveness of quantity demanded to a change in price‟.

2. Define the term “cost”. Cost may be defined as a total of all expenses incurred, whether paid or outstanding, in the manufacture and sale of a product.

3. What is opportunity cost?

Opportunity cost may be defined as the potential benefit that is given up as an alternative course of action is sought. In other words, the expected return of benefit for gone in rejecting one course of action for another.

4. What do you mean by marginal cost? The Institute of Costs and Works Accountants of India defined marginal cost as, “the amount at any given volume of output by which aggregate costs are changed, if the volume of output is increased or decreased by one unit”.

5. Explain marginal costing? Marginal costing is defined by the ICWA as, “the ascertainment by differentiating between fixed costs, and variable costs, of marginal costs and of the effect on profit of changes in volume or type of output”.

6. What is meant by marginal revenue?

The revenue that can be obtained from selling one more unit of product is called marginal revenue.

7. Give a short note on sunk cost. (M/J 2012) A cost which was incurred or sunk in the past and is not relevant to the particular

decision-making is a sunk cost or sunk loss. It may be variable or fixed or both.

8. List out the elements of cost.

The elements of cost are a. Materials b. Labour and c. Expenses.

9. Define the term costing. Institute of Costs and Management Accountants (ICMA), London, has defined costing as the ascertainment costs. “It refers to the techniques and processes of ascertaining costs and studies the principles and rules concerning the determination of cost of products and services”.

10. What is breakeven point? The breakeven point is that volume of output at which neither a profit is made nor a loss is incurred. It is a point where the total sales are equal to total cost.

11. Write the formula to find P/V ratio. The formula for computing the P/V ratio is given below.

Contribution

P/V ratio−−−−−−−−−−=

Sales Fixed cost + Profit

−−−−−−−−−−−−−−Or= Sales

−VSariableles

cost

−−−−−−−−−−−−−−−−Or= Sales

12. List out the formulae to find out the margin of safety.

Margin−Sales of safetyB.E=.PActual sales Profit

Or−−−−−−−= P/V ratio

Profit

Or−−−−−−−−−−=×Sales Contribution Margin of safety

As −−−−−−−−−−−−−a percentage×100= Total sales

13. What is fixed cost? Fixed cost means that the cost tends to be unaffected with the volume of output. Fixed cost depends

upon the passage of time and does not vary directly with the volume of output. It is known as period

cost, (e.g) rent and rates of factory buildings, insurance of buildings, depreciation of buildings, etc.

14. Give a short note on variable cost. Variable cost tends to vary directly with the volume of output. It varies almost in direct proportion to the volume of production, (e.g) the cost of direct materials, direct labour, direct chargeable expenses such as power, repairs, etc, If production increases, the costs will also increase and vice versa.

15. What is contribution? Contribution is the difference between sales and marginal cost of sales. The formulae for contribution are as follows:

−Contributional cost= Sales

Or − =Variable

Sales cost Or = Fixed cost ± Profit or loss

UNIT – II: VALUE ENGINEERING

1. What do you mean by „Make or Buy Decisions‟?

Make or buy decision is a determination whether to produce a component part internally or to buy it from an outside supplier. The organization should evaluate the costs and benefits of manufacturing a product or product component against purchasing it and then select the alternative which results in the lower cost.

2. What are the different approaches followed in make or buy decisions?

The following are the approaches followed in make or buy decisions. a. Simple cost analysis b. Economic analysis c. Break-even analysis.

3. What is meant by value analysis/value engineering?

Value analysis is a special type of cost reduction technique. It critically investigates and analyses the different aspects of materials, design, cost and production of each and every component of the product in order to produce it economically without decreasing its utility, function or reliability.

4. What do you understand by value of a product?

Value differs from both price and cost in the sense that it is the cost proportionate to the function. We can express value mathematically as

Function or utility −−−−−−−−−−−−−−Value=

Cost

5. Explain „function‟.

Function specifies the purpose of the product or what the product does, what is its utility etc.

6. What are the various functions of a product?

Functions can be classified into the following three categories: a. Primary functions b. Secondary functions c. Tertiary functions.

7. What are the various types of values?

a. Cost value b. Exchange value c. Use value d. Esteem value.

8. Write any four objectives of value analysis.

b. Simplify the product c. Use (new) cheaper and better materials d. Modify and improve product design so as to make it acceptable to consumer.

9. Differentiate value analysis and value engineering.

Sl.no Value analysis Value engineering

1 Value analysis is the application of a set Value engineering is the application of of techniques to an existing product exactly the same set of techniques to a

with a view to improve its value. new product at design stage.

2 It is a remedial process. It is a preventive process.

10. List any four advantages of value engineering.

a. Value engineering/analysis identify and reduce the product cost b, It modifies and improves the product design c. It increases the performance/utility of the product by economical means d. It helps to generate new ideas.

11. What is sinking fund factor?

The factor i/(1+i)N −1 is called the equal payment series sinking fund factor or sinking fund factor, and is referred to by the notation (A/F, i, N). A sinking fund is an interest-bearing account into which a fixed sum is deposited each interest period; it is commonly established for the purpose of replacing fixed assets.

12. What is capital recovery factor?

The factor i(1+i)N / (1+i)N −1 is called the equal payment series capital recovery f actor, or

simply capital recovery factor, which is designated (A/P, i, N). In finance this A/P factor is referred to as the annuity factor. The annuity factor indicates a series of payments of a fixed, or constant, amount for a specified number of periods.

!3. What is uniform gradient conversion?

(1+ i)N −i−1N −−−−−−−−−−−−The factor is called the uniform factor or the gradient to equal payment

i[(1+i)N −1] series conversion factor and is designated (A/G, i , N). The future sum of annual equal payments at the end of every year for N years is equal to the total amount of gradient series at the end of N years.

14. What is „effective interest rate‟?

Effective interest rate is a percentage that is periodically applied to measure the cost of money when the interest rates compounded for less than a year i.e., monthly, quarterly and half yearly. The formula to compute effective interest rate isR = (1+i/C)C −1

Wherei=thenominalinterestrate, C=thenumberofinterestperiodsinayear.

15.ExplaintheconceptofDiscounting. Findingthepresentworthofafuturesumissimplythereverseofcompoundingand isknownasdiscountingprocess.Theformulatoobtainpresentworthis P=F(P/F,i,N)

where(P/F,i,N)isthefactornotationforsinglepaymentpresentworthfactor.The interestrateiandP/Ffactorarealsoreferredtoasthediscountrateand discountingfactor, respectively.

UNIT – III: CASH FLOW

1. What is revenue dominated cash flow?

The profit/revenue, salvage value of all inflows to an organization will be assigned with positive sign and the cost outflows will be assigned with negative sign and this is called revenue dominated cash flow.

2. What is cash dominated cash flow?

The cost outflows will be assigned with positive sign and the profit, revenue, salvage value,

all inflows etc., will be assigned with negative sign and this is called cost dominated cash

flow.

3. What is annual equivalent method of comparing alternatives?

In the case of revenue dominated cash flow, the corresponding annual equivalent revenues are to be computed and compared, then the alternative with the maximum annual equivalent revenue should be selected as the best alternative. In case of cost dominated cash flow the corresponding annual equivalent costs are to be computed and compared, then the alternative with the minimum annual equivalent cost should be selected as the best alternative.

4. Mention the various rate of return

methods. The various rate of return

methods are:

a. Internal rate of return (IRR) b. Average rate of return (ARR) c. Net preset value method (NPV) d. Pay-back period (PBP).

5. What is rate of return?

Rate of return is the break-even interest rate, i, which equates the present worth of a project‟s cash outflows to the present worth of its cash inflows.

6. What is present worth method?

The present worth measures the surplus in an investment project at time zero (0). The present worth of all cash inflows is computed against the present worth of all cash outflows associated with an investment of project is called present worth method.

7. What is future worth analysis?

Net future worth measures the surplus at a time period other than 0. Future worth analysis is particularly useful in an investment situation where we need to compute the equivalent with of a project at the end of its investment period.

8. What is annual equivalent method?

The criterion provides a basis for measuring investment worth by determining equal payments on an annual basis is called annual equivalent method.

9. Write down the techniques for comparing the worthiness of a project.

a. Net present value methods b. Rate of return methods c. Ratio methods d. Pay back methods e. Annual equivalent methods f. Future worth method.

10. Define IRR.

IRR (Internal Rate of the Return) is the rate of return at which total present value of Future cash inflow is equal to initial investment. The rate of return is generally found by trial and error method.

11. Define MARR.

MARR (Minimum Attractive Rate of Return) represents the required or minimum interval rate of return for a project. The MARR is a minimum return the company will accept on the money it invests.

12. What are the advantages of Rate of Return

method? The advantages of Rate of Return method

are:

a. It is easy to understand and operate b. It uses the entire earnings of an investment proposal, unlike the payback period method. c. It gives a clear picture of the profitability of a project.

13. What are the disadvantages of Rate of Return

method? The disadvantages of Rate of Return

method are:

a. Ignores the time value of the money and treats all incomes same irrespective of their time of receipt.

b. Reliability is affected due to the various concepts of investment. c. Cash flow aspect of projects is not properly assessed. d. Not useful to assess projects where investment is made in two or more installments at

different times.

14. What is the time value of money?

The economic value of a sum depends on when it is received. Because money has earning power over time, (it can be put to work, earning more money for its owner), a rupee received today has a greater value than a rupee received at some future time.

UNIT – IV: REPLACEMENT AND MAINTENANCE ANALYSIS

1. What is replacement analysis? Replacement analysis involves the replacement of existing obsolete or worn-out assets in order to avoid failure in operations. The problems often faced by management of various industries are whether to replace the existing equipment with new and more efficient equipment or to continue to use existing equipment, and when existing equipment should be replaced with efficient equipment. This class of decision analysis is known as replacement analysis.

2. What is meant by gradual failure?

Gradual failure is progressive in nature. It can be depicted in machine equipment, which is gradually depreciating or deteriorating with the time resulting in very high operating and maintenance costs and/or decreased residual value. It is easier to predict such type of failures and take the corrective measures by providing a replacement policy for such machine equipment.

3. Define economic service life of an asset. The economic service life of an asset is defined to be the period of useful life that minimizes the annual equivalent cost of owning and operating the asset.

4. What are the types of replacement problem? a. Replacement of assets that deteriorate with time (replacement due to gradual failure, or

wear and tear of the components of the machine). This can be further classified into the following types: i) Determination of economic life of an asset, ii) Replacement of an asset with a new asset.

b. Simple probabilistic model for assets that fails completely. (Replacement due to sudden failure)

5. Explain capital recovery cost. Capital recovery cost computed from the first cost (Initial investment/purchase price) of the machine. Generally speaking, as an asset becomes older, its salvage value becomes smaller. As long as The salvage value is less than the initial cost, the capital recovery cost is a decreasing function of the life of the asset. In other words, the longer we keep an asset, the lower the capital recovery cost becomes.

6. Explain operating costs. The operating costs of an asset include operation and maintenance (O&M) costs, labour costs,

material costs, and energy consumption costs. O&M costs tend to increase as a function of the age

of the asset. Because of increasing trend of the O&M costs, the total operating costs of an asset

usually increases as the asset ages. As long as the annual operating cost increases with the age of

the equipment, the annual equivalent operating cost is an increasing function of the life of the asset.

7. Explain annual equivalent total cost. Annual equivalent total cost of owning and operating an asset is a summation of the capital

recovery cost (average first cost) and the annual equivalent operating costs of the asset.

[

8. Explain sunk costs.

The purchase costs of an equipment three years ago and repair cost of last year are called as

sunk costs. A sunk cost is any past cost unaffected by any future investment decision. In a proper engineering economic analysis, only future costs should be considered; past sunk costs should be ignored.

9. What is meant by maintenance?

Maintenance is concerned with the day-to-day problem of keeping production facilities and equipment in proper operating condition. The machines and equipments should be continuously monitored for this efficient functioning. Otherwise, the quality of service will be poor and the cost of operation and maintenance would increase with the passage of time.

10. Name the types of maintenance. a. Corrective or Breakdown maintenance b. Scheduled maintenance c. Preventive maintenance d. Predictive maintenance.

11. State the main causes of breakdown.

a. Failure to replace worn out parts b. Lack of lubrication c. Indifference towards minor faults.

12. State any two disadvantages of breakdown maintenance.

a. Delays in production b. Faster plant deterioration.

13. What is predictive maintenance?

In predictive maintenance, equipment conditions are monitored and measured periodically or

on continuous basis and this enables maintenance man to take action such as equipment

adjustment, repair or overhaul. This will extend the service life of equipment without fear of

failure.

14. What is the significance of timely maintenance? Timely maintenance makes more running time of machines and equipments and helps in the

continuous production thereby improving the productivity of the organisation.

15. What is preventive maintenance? Preventive maintenance is defined as any action performed in an attempt to keep a machine in

a specified operating condition by means of systematic inspection, detection, and prevention of

incipient failures. It minimizes breakdown maintenance.

UNIT – V: DEPRECIATION

1. Define the term “Depreciation”.

Depreciation is the process of allocating the acquisition cost of the tangible assets less salvage value, if any, in a systematic and a rational manner over the estimated life of the asset.

2. Mention the various methods used in depreciation calculation. The

various methods used in depreciation calculations are:

a. Straight line method b. Declining balance method c. Sum of the years digits method d. Sinking fund of annuity method e. Service output method.

3. What are the causes of depreciation? The

causes of depreciation are:

a. Wear and tear b. Depletion c. Obsolescence d. Lapse of time.

4. Write five reasons for providing depreciation.

The reasons for providing depreciations are:

a. To know the correct profits b. To show correct financial position c. To make provision for replacement of assets d. To compute tax liability e. To decide for how much to buy or sell the assets in the second-hand market.

5. How to compute the sum of the digits of the years, if an asset has a life of six years?

Sum of the years = 1+2+3+4+5+6 = 21 n (n+1)

21 = −−−−−−− 2

6 (6+1) −−−−−−−=

2 = 42/2 = 21

6. What is evaluation of public alternatives?

Evaluation of public alternative is nothing but the selecting of best alternative from the available alternatives.

7. What is the main objective of evaluation of public alternatives? To provide goods and services to the public at the minimum cost is the main objective of evaluation of public alternatives. In this situation, public alternative evaluation must consider a point that whether the benefits of the public activity are at least equal to its costs of consumption during the job.

8. What is „Book value‟?

The value at which an asset is carried on a balance sheet is called book value. In other words the cost

of an asset minus accumulated depreciation is known as book value.

9. Define the term “Benefit cost ratio”

The ratio between the equivalent benefit and equivalent costs is called the benefit cost ratio. Equivalent benefits

i.e BC−−−−−−−−−−−−−−ratio=

Equivalent costs

10. Define the term “inflation‟

Inflation may be defined as a sustained rise in the general price level. It is an economic condition

where there is a rise in prices resulting in the fall in the purchasing power of money.

11. What are the types of inflation? The

types of inflation are: a. Creeping inflation b. Moderate inflation c. Galloping inflation d. Hyper inflation.

12. What is service output method of depreciation?

Service output method of depreciation is a type of computing depreciation based on service rendered by an asset.

13. What is sinking fund? A depreciation fund, equal to be actual loss in the value of the asset, is estimated for each year. This amount is invested outside the business in a separate account called sinking fund investment account at a certain rate and the interest will be earned on the fund. Therefore the sinking fund will rise year after year.

14. What is amortization?

Amortization is a routine decrease in value of an intangible asset, or the process of paying off a debt over time through regular payments. Amortization refers to the expensing of intangible capital assets

(intellectual property: patents, trademarks, copyrights, etc.) in order to show their decrease in value as

a result of use or passage of time.

15. What is functional depreciation? Functional depreciation occurs as a result of changes in the organisation or in technology that decrease or eliminate the need for an asset. Examples of functional depreciation include obsolescence attributable to advances in technology, a declining need for the services performed by an asset, or the inability to meet increased quantity and/or quality demands.

PART – B (15 Marks)

UNIT I - INTRODUCTION TO ECONOMICS

1.1 Introduction

Efficient functioning of any business organization would enable it to provide goods/services at a lower price.

In the process of managing organizations, the managers at different levels should take appropriate economic decisions which will help in minimizing investment, operating and

maintenance expenditures besides increasing the revenue, savings and such other gains of the organization.

These can be achieved through Engineering Economics which deals with the methods that enable one to make economic decisions towards minimizing c sts and/or maximizing

benefits to business organizations.

This chapter discussesCivildatastheelementsofeconomics and the interaction

between its various components.

This is followed by an analysis of the need nd scope of engineering economics. Later, elements of cost and break-even analysis are presen ed.

1.2 Economics - is a study of economic problems of the people concerning production,

consumption, exchange and distribution of wealth.

Economics is the science that dea with the production and consumption of goods and

services and the distribution and rendering of these for human welfare.

The following are the economic goals.

✓ A high level of employment

✓ Price stability

✓ Efficiency

✓ An equitable distribution of income

✓ Growth

1.3 Flow in economy The flow of goods, services, resources and money payments in a simple

economy

Households and businesses are the two major entities in a simple economy.

Business organizations use various economic resources like land, labour and

capital hich are provided by households to produce consumer goods and services

which ill be used by them.

Business organizations make payment of money to the households for receiving various resources.

The households in turn make payment of money to business organizations for receiving consumer goods and services.

This cycle shows the interdependence between the two major entities in

a simple economy.

Money payments for consumer goods and services

Consumer goods, services

Businesses

Households com

1. Consume final g ods

1. Provide goods and services to

and services produced

consumers

by businesses and services

2. Use resources, inputs

2. Provide pr du tive

provided by households

inputs to businesses

Civildatas

Money payments for resources, rent , wage ,

salaries, interest and profit

Economic Resources: Land, l bour, c pit l Fig. Flow of goods, services, resources and money p yments in a simple economy.

1.4 Laws of supply and demand

1.3.1 Laws of supply - states that the quantity of commodity supplied varies directly

with the price, other determinants of supp y remaining constant.

✓ If the cost of inputs increases, then naturally, the cost of the product will go up. In such a situation, at the pre ailing price of the product the profit margin per unit will

be less. ✓ The producers w ll then reduce the production quantity, which in turn will

affect the supply of the product. ✓ For instance, if the prices of fertilizers and cost of labour are

increased significantly, in agriculture, the profit margin per bag of paddy will be reduced.

✓ So, the farmers will reduce the area of cultivation, and hence the quantity of supply of paddy will be reduced at the prevailing prices of the paddy.

✓ If there is an advancement in technology used in the manufacture of the product in the long run, there will be a reduction in the production cost per unit.

✓ This ill enable the manufacturer to have a greater profit margin per unit at the

prevailing price of the product. Hence, the producer will be tempted to supply more quantity to the market.

✓ Weather also has a direct bearing on the supply of products. For example, demand for woollen products will increase during winter. This means the prices

of woollen goods will be incresed in winter. ✓ So, naturally, manufacturers will supply more volume of woollen goods

during winter.

Factors influencing supply

The shape of the supply curve is affected by the following factors:

Cost of the inputs

Technology

Weather

Prices of related goods

1.3.2 Law of demand states that other things being equal demand when price falls

and contracts when price rises.

Market demand is the total quantity demanded by all the purchasers together.

9. Elasticity of Demand - Elasticity of demand may be defined as the degree of responsiveness of quantity demanded to a Change in pri e.

9. An interesting aspect of the economy is that the demand and supply of a product

are interdependent and they are sensitive with respect to the price of that

product. .

10. From Fig. it is clear that when there is decre se in the price of a product, the demand for the product increases and i s supply decreases.

11. Also, the product is more in dem nd nd hence the demand of the product

increases.

12. At the same time, lowering of the price of the product makes the producers restrain from releasing more quantities of the product in the market.

13. Hence, the supply of the product is decreased. The point of intersection of the supply curve and the demand curve is known as the equilibrium point.

14. At the price corresponding to this point, the quantity of supply is equal to the quantity of demand. Hence, this point is called the equilibrium point.

Factors influencing demand

The shape of the demand curve is influenced by the following factors:

Income of the people

Prices of related goods

Tastes of consumers

1.5 Concept of Engineering Economics

Science is a field of study where the basic principles of different physical systems are formulated and tested.

Engineering is the application of science. It establishes varied application systems based on different scientific principles.

14. From the discussions in the previous section, it is clear that price has a major role in deciding the demand and supply of a product.

15. Hence, from the organization’s point of view, efficient and effective functioning of

the organization would certainly help it o provide goods/services at a lower cost

which in turn will enable it to fix a lower price for its goods or services.

16. The following section discusses the different types of efficiency and their impact on

the operation of businesses and the definition and scope of engineering economics. . 1.5 Types of Efficiency

Efficiency of a system s generally defined as the ratio of its output to input. The efficiency can be class f ed into technical efficiency and economic efficiency.

1. Technical efficiency

It is the ratio of the output to input of a physical system. The physical system

may be a diesel engine, a machine working in a shop floor, a furnace, etc.

Technical efficiency (%) = Output

× 100

Input

The technical efficiency of a diesel engine is as follows:

Heat equivalent of mechanical

Technical efficiency (%) = energy produced

× 100

Heat equivalent of fuel used

In practice, technical efficiency can never be more than 100%.

This is mainly due to frictional loss and incomplete combustion of fuel, which are considered to be unavoidable phenomena in the working of a diesel engine.

2. Economic efficiency

Economic efficiency is the ratio of output to input of a business system.

Economic efficiency (%) = Output

× 100 = Worth

× 100

Input Cost

‘Worth’ is the annual revenue generated by way f perating the business

and ‘cost’ is the total annual expenses incurred in carrying ut the business.

For the survival and growth of any business, the economic efficiency should be more than 100%.

Economic efficiency is also called ‘productivity’. There are several ways

of improving productivity.

13. Increased output for the s me input

14. Decreased input for the s me output

15. By a proportionate increase in the output which is more than the proportionate increase in the input

16. By a proport onate decrease in the input which is more than the proportionate decrease in the output

17. Through simultaneous increase in the output with decrease in

the input.

1.6 Definition and Scope of Engineering Economics

As stated earlier, efficient functioning of any business organization would enable it to provide goods/services at a lower price.

In the process of managing organizations, the managers at different levels should take appropriate economic decisions which will help in

minimizing investment, operating and maintenance expenditures besides increasing the revenue, savings and other related gains of the organization.

1.6.1 Definition

Engineering economics deals with the methods that enable one to take economic decisions towards minimizing costs and/or maximizing benefits to

business organizations.

1.6.2 Scope

The issues that are covered in this book are elementary economic analysis,

interest formulae, bases for comparing alternatives, present worth method, future

worth method, annual equivalent method, rate of return method, replacement

analysis, depreciation, evaluation of public alternatives, inflation adjusted

investment decisions, make or buy decisions, inventory control, project

management, value engineering, and linear programming.

1.7 Elements Of Costs

Cost can be broadly classified into variable c st and overhead cost. Variable cost varies with the volume of production while verhead cost is fixed, irrespective

of the production volume.

Variable cost can be further classified into direct material cost, direct labour

cost, and direct expenses. The overhead co t can be classified into factory overhead,

administration overhead, selling overhead, and distribution overhead.

Direct material costs are those cos s of materials that are used to produce

the product. Direct labour cost is the mount of wages paid to the direct labour

involved in the production activities.

Direct expenses are those expenses that vary in relation to the production volume, other than the d rect material costs and direct labour costs.

Overhead cost is the aggregate of indirect material costs, indirect labour costs

and indirect expenses. Administration overhead includes all the costs that are incurred in admin ster ng the business.

Selling overhead is the total expense that is incurred in the promotional

activities and the expenses relating to sales force. Distribution overhead is the total cost of shipping the items from the factory site to the customer sites.

The selling price of a product is derived as shown below:

14. Direct material costs + Direct labour costs + Direct expenses = Prime cost

Prime cost + Factory overhead = Factory cost

15. Factory cost + Office and administrative overhead

Costs of production

16. Cost of production + Opening finished stock – Closing finished

stock

Cost of goods sold

16. Cost of goods sold + Selling and distribution overhead = Cost of sales

17. Cost of sales + Profit = Sales

18. Sales/Quantity sold = Selling price per unit

In the above calculations, if the opening finished stock is equal to the closing finished stock, then the cost of production is equal to the cost of goods sold.

1..8 Other Costs/Revenues

The following are the costs/revenues other than the costs which are presented in the previous section:

Marginal cost

Marginal revenue

Sunk cost

Opportunity cost

1. Marginal Cost

Marginal cost of a product is the cost of producing an additional unit of that

product. Let the cost of producing 20 units of a product be Rs. 10,000, and the cost of producing 21 units of the same product be Rs. 10,045. Then the marginal cost of producing

the 21st unit is Rs. 45.

2. Marginal Revenue

Marginal revenue of a product is the incremental revenue of selling an additional unit of that product Let, the revenue of selling 20 units of a product be Rs. 15,000 and the

revenue of selling 21 units of the same product be Rs. 15,085. Then, the marginal revenue of selling the 21st unit is Rs. 85.

3. Sunk Cost

This is known as the past cost of an equipment/asset. Let us assume that an equipment has been purchased for Rs. 1,00,000 about three years back. If it is considered

for replacement, then its present value is not Rs. 1,00,000. Instead, its present market value should be taken as the present value of the equipment for further analysis.

So, the purchase value of the equipment in the past is known as its sunk cost.

The sunk cost should not be considered for any analysis done from now onwards.

4. Opportunity Cost

In practice, if an alternative (X ) is selected from a set of competing alternatives

(X,Y ), then the corresponding investment in the selected alternative is not available for any

other purpose. If the same money is invested in some other alternative (Y ), it may fetch

some return. Since the money is invested in the selected alternative (X ), one has to forego

the return from the other alternative (Y ).

The amount that is foregone by not investing in the other alternative (Y ) is known as the opportunity cost of the selected alternative (X ). So the opportunity cost of an

alternative is the return that will be foregone by not investing the sa e oney in another alternative.

1.9 Break-Even Analysis

The main objective of break-even analysis is to find the ut-off production volume from

where a firm will makeCivildatasprofit.

Let

s = selling price per unit

v = variable cost per unit

FC = fixed cost per period

Q = volume of production

The total sales revenue (S) of the firm is given by the following formula:

S = s Q

The total cost of the f rm for a given production volume is given as

TC = Total variable cost + Fixed

cost

= v Q + FC

• The linear plot of the above two equations are shown in Fig. .

• The intersection point of the total sales revenue line and the total cost line is called the break-even point.

• The corresponding volume of production on the X-axis is known as the break-even sales quantity.

• At the intersection point, the total cost is equal to the total revenue.

This point is also called the no-loss or no-gain situation.

• For any production quantity which is less than the break-even quantity, the total cost is more than the total revenue.

• Hence, the firm will be making loss.

Sales (S)

Profit Total cost (TC)

Break-even Variable cost (VC) sales Fixed cost (FC)

Loss

BEP(Q*) Production quantity Fig. Break-even chart.

For any production quantity which is more than the break-even quantity, the Civildatas

total revenue will be more than the total cost Hence, the firm will be making profit.

Profit = Sales – (Fixed co t + Variable costs)

= s Q – (FC + v Q)

The formulae to find the break-even qu ntity and break-even sales quantity

Break-even quantity = Fixed cost

Selling

price/unit − Variable cost/unit

FC

= s-v (in units)

Break-even sales = Fixed cost × Selling price/ unit

Selling price/ unit − Variable cost / unit

= FC

× (Rs.)

s − v

The contribution is the difference between the sales and the variable costs. The

margin of safety (M.S.) is the sales over and above the break-even sales. The formulae to compute these values are

Contribution = Sales – Variable costs

Contribution/unit = Selling price/unit – Variable cost/unit

M.S. = Actual sales – Break-even sales

= Profit

× sales

Contribution

M.S. as a per cent of sales = (M.S./Sales) 100

1.10 Profit/Volume Ratio (P/V Ratio)

P/V ratio is a valid ratio which is useful for further analysis.

The different formulae for the P/V ratio are as follows:

Contribution Sales − Variable costs

P/V ratio = Sales = Sales

The relationship between BEP and P/V ratio is as follows:

Fixed cost BEP =

1.11 Elementary Economic Analysis

P/V ratio

Whether it is a business situation or a day-to-day event in somebody’s personal

life, there are a large number of economic decision making involved. One can

manage many of these decision problems by using simple economic analysis.

For example, an industry can source its r w m eri ls from a nearby place or from

a far-off place. In this problem, the following fac ors will affect the decision:

Price of the raw m teri l

Transportation cost of the raw material

`Availability of the raw material

Quality of the raw material

Consider the alternati e of sourcing raw materials from a nearby

place with the follow ng characteristics:

The raw material is more costly in the nearby area.

The availability of the raw material is not sufficient enough

to support the operation of the industry throughout the year.

✓ The raw material requires pre-processing before it is used in

the production process. This would certainly add cost to the

product.

✓ The cost of transportation is minimal under this alternative.

On the other hand, consider another alternative of sourcing the raw

materials from a far-off place with the following characteristics:

The raw material is less costly at the far off place.

The cost of transportation is very high.

The availability of the raw material at this site is abundant

and it can support the plant throughout the year.

The raw material from this site does not require any pre-

processing before using it for production.

1.12 Material Selection For A Product/Substitution Of Raw Material

The cost of a product can be reduced greatly by substitution of the raw materials.

Among various elements of cost, raw material cost is most significant and it forms a major portion of the total cost of any product.

So, any attempt to find a suitable raw material will bring a reduction in the total cost in any one or combinations of the following ways:

Reduced machining/process time

Enhanced durability of the product

Cheaper raw material price

Therefore, the process of raw material selection/substitution will result in

finding an alternate raw material which will provide the nece ary functions that are provided by the raw material that is presently used.

In this process, if the new raw materi l provides any additional benefit, then it should be treated as its welcoming feature. This concept is demonstrated with

numerical problem given below

Example

In the design of a jet eng ne part, the designer has a choice of specifying either an aluminium alloy casting or a steel cast ng. Either material will provide

equal service, but the aluminium casting will weigh 1.2 kg as compared with 1.35 kg

for the steel casting.

The aluminium can be cast for Rs. 80.00 per kg. and the steel one for Rs. 35.00

per kg. The cost of machining per unit is Rs. 150.00 for aluminium and Rs. 170.00

for steel. Every kilogram of excess weight is associated with a penalty of Rs. 1,300

due to increased fuel consumption. Which material should be specified and what is

the economic advantage of the selection per unit?

Solution (a) Cost of using aluminium metal for the jet

engine part:

Weight of aluminium casting/unit = 1.2 kg Cost of making aluminium casting = Rs. 80.00 per kg Cost of machining aluminium casting per unit = Rs. 150.00 Total cost of jet engine part made of aluminium/unit

5. Cost of making aluminium casting/unit

a Cost of machining aluminium casting/unit

6. 80 1.2 + 150 = 96 + 150 7. Rs. 246

9. Cost of jet engine part made of

steel/unit: Weight of steel casting/unit =

1.35 kg Cost of making steel casting = Rs. 35.00 per kg Cost of machining steel casting per unit = Rs. 170.00

Penalty of excess weight of steel casting = Rs. 1,300 per

kg

Total cost of jet engine part made of steel/unit 10. Cost of making steel casting/unit

Cost of machining steel casting/unit

Penalty for excess weight of steel casting

11. 35 1.35 + 170 + 1,300(1.35 – 1.2) 12. Rs. 412.25

DECISION The total cost/unit of a jet engine part made of

aluminium is less than that for an engine made of steel. Hence,

aluminium is suggested for making the jet engine part. The

economic advantage of using aluminium over steel/unit is Rs.

412.25 – Rs. 246 = Rs. 166.25

1.12.1 Design Selection for a Pro uct

a. The design modification of a product may result in reduced raw material

requirements, ncreased machinability of the materials and reduced

labour.

12. Design is an mportant factor which decides the cost of the product for a specified level of performance of that product.

Example

(Design selection for a process industry). The chief engineer of refinery

operations is not satisfied with the preliminary design for storage tanks to

be used as part of a plant expansion programme. The engineer who

submitted the design was called in and asked to reconsider the overall dimensions in

the light of an article in the Chemical Engineer, entitled “How to size

future process vessels?” The original design submitted called for 4 tanks 5.2 m in diameter and 7

m in height. From a graph of the article, the engineer found that the present

ratio of height to diameter of 1.35 is 111% of the minimum cost and that the minimum cost for a tank was when the ratio of height to diameter was 4

: 1. The cost for the tank design as originally submitted was estimated to be Rs. 9,00,000. What are the optimum tank dimensions if the volume remains

the same as for the original design? What total savings may be expected through the redesign?

Solution

13. Original design Number

of tanks = 4 Diameter of

the tank = 5.2 m Radius of

the tank = 2.6 m Height of

the tank = 7 m

Ratio of height to diameter = 7/5.2 = 1.35

Volume/tank = (22/7)r 2

h = (22/7)(2.6)2

7

148.72 m3

15. New design Cost of the old design = 111% of the cost of the new design (optimal

design) Optimal ratio of the height to diameter = 4:1

h : d = 4 : 1

4d = h

d = h/4

r = h/8

Volume = (22/7)r2

h = 148.72 (since, he volume remains the same)

(22/7)(h/8)2

h = 148.72 3

148.72

h = 64 = 3,028.48

h = 14.47 m

r = h/8 = 14.47/8 = 1.81 m

Therefore, Diameter of the new design = 1.81 2

= 3.62 m

Cost of the new design = 9,00,000 (100/111)

= Rs. 8,10,810.81

Expected savings by the redesign = Rs. 9,00,000 – Rs.

8,10,810.81 = Rs. 89,189.19

1.13 Process Planning /Process Modification

While planning for a new component, a feasible sequence of operations with the least cost of processing is to be considered.

The process sequence of a component which has been planned in the past is not static.

It is always subject to modification with a view to minimize the cost of manufacturing the component.

So, the objective of process planning/process modification is to identify the most economical sequence of operations to produce a c mp nent.

The steps in process planning are as follows:

1. Analyze the part drawing to get an overall picture of what is required.

2. Make recommendations to or consult with product engineers on

product design changes. 3. List the basic operations required o produce the part to

the drawing or specific tions.

4. Determine the most pr ctic l nd economical manufacturing method and the form or tooling required for each operation.

5. Devise the best way to combine the operations and put them in

sequence. 6. Specify the gaug ng required for the process.

Steps 3–5 a m to determine the most practical and economical sequence

of operations to produce a component. This concept is demonstrated with a

numerical problem.

Example

The process planning engineer of a firm listed the sequences of operations as shown in Table to produce a component.

Table Data for Example

Sequence Process sequence

1 Turning – Milling – Shaping – Drilling 2 Turning – Milling – Drilling 3 All operations are performed with CNC machine

The details of processing times of the component for various

operations and their machine hour rates are summarized in Table below

Machine Hour Rates and Processing Times (minutes) for Example

Operation Machine hour Process

sequence rate

(Rs.) 1 2 3

Turning 200 5 5

– Milling 400 8 14

– Shaping 350 10 –

– Drilling 300 3 3

– CNC operations 1,000 – – 8

Find the most economical sequence of operations t manufacture the

component.

Solution (a) Cost of component using process sequence 1. The process

sequence 1 of the component is as follows:

Turning – Milling – Shaping – Drilling

The calculations for the cost of the above process sequence are summarized in Table.

Table Workings for Process Sequence 1

Operation Operation Time Machine Cost

No. hour rate

(min) (hr) (Rs.) (Rs.)

1 Turning 5 0.083 200 16.60

2 Milling 8 0.133 400 53.20

3 Shaping 10 0.167 350 58.45

4 Dr ll ng 3 0.050 300 15.00

Total: 143.25

(b) ost of component using process sequence 2. The process

sequence 2 of the component is as follows:

Turning – Milling – Drilling

The calculations for the cost of the above process sequence are given in

Table

Table Workings for Process Sequence 2

Operation Operation Time Machine Cost No. hour rate

(min) (hr) (Rs.) (Rs.)

1 Turning 5 0.083 200 16.60 2 Milling 14 0.233 400 93.20 3 Drilling 3 0.050 300 15.00

Total: 124.80

2. Cost of component using process sequence 3. The process sequence 3 of the component is as follows:

Only CNC operations

The calculations for the cost of the above process sequence are

summarized in Table

Table Workings for Process Sequence 3

Operation Operation Time Machine Cost

No. hour rate

(min) (hr) (Rs.) (Rs.)

1 CNC operations 8 0.133 1,000 133

The process sequence 2 has the least cost. Therefore, it sh uld be selected

for manufacturing the component.

UNIT-II VALUE ENGINEERING

2.1 INTRODUCTION

Value analysis is one of the major techniques of cost reduction and cost

prevention. It is a disciplined approach that ensures necessary functions for minimum cost without sacrificing quality, reliability, performance, and appearance. According

to the Society of American Value Engineers (SAVE),

Value Analysis is the systematic application ofcomrecognized techniques which

identify the function of a product or service, establish a onetary value for the function

and provide the necessary function reliably at the lowest overall cost.

It is an organized approach to identify unnecessary sts associated with any

product, material part, component, system or service by analysing the function and eliminating such costs without impairing the quality, fun tional reliability, or the

capacity of the product to give service

2.1.1 WHEN TO APPLY VALUE ANALYSIS

One can definitely expect very good resul s by initiating a VA programme if one or more of the following symp oms are present:

3. Company’s products show decline in sales.

4. Company’s prices are higher than those of its competitors.

5. Raw materials cost has grown disproportionate to the volume of production.

6. New designs are be ng introduced.

7. The cost of manufacture is rising disproportionate to the

volume of production.

4. Rate of return on investment has a falling trend.

5. Inability of the firm to meet its delivery commitments.

2.1.2 Value

The term ‘value’ is used in different ways and, consequently, has different

meanings. The designer equates the value with reliability; a purchase person with

price paid for the item; a production person with what it costs to manufacture, and a

sales person with what the customer is willing to pay.

Example

Cost value. It is the summation of the labour, material, overhead and all other

elements of cost required to produce an item or provide a service compared to a base.

Exchange value. It is the measure of all the properties, qualities and features of

the product, which make the product possible of being traded for another product or

for money.

Value analysis/value engineering

It is a special type of cost reduction technique. It critically investigates and

analyses the different aspects of materials, design, cost and production of each and

every component of the product in produce it economically without decreasing its utility, function or reliability.

Applications

The various application areas of value engineering are machine tool

industries, industries making accessories for machine tools, auto industries, import substitutes, etc.

2.2 INTRODUCTION

In the process of carrying out business a tivities of an organization, a

component/productcanbemadewithintheorganization or bought from a subcontractor.

Each decision involves its own costs.

So, in a given situation, the organization should evaluate each of the above make or buy alternatives and then select the alterna ive which results in the lowest cost. This is an important decision since it affects the produc ivi y of the organization

In the long run, the make or buy ecision is not static. The make option of a

component/product may be economical to ay; but after some time, it may turn out to be

uneconomical to make the same. Thus, the make or buy dec on should be reviewed periodically, say, every 1 to 3

years. This is mainly to cope with the changes in the level of competition and various other environmental factors.

Make or Buy Decisions - is determination whether to produce a component part

internally or to buy it from an outside supplier. The Organization should evaluate the

costs and benefits of manufacturing a product or product component against purchasing it

and then select the alternative which results in the lower cost.

2.2.1 CRITERIA FOR MAKE OR BUY

In this section the criteria for make or buy are discussed.

7. Criteria for make

The following are the criteria for make:

a. The finished product can be made cheaper by the firm than by outside suppliers.

The finished product is being manufactured only by a limited

number of outside firms which are unable to meet the demand.

The part has an importance for the firm and requires extremely close quality control.

The part can be manufactured with the firm’s existing facilities

and similar to other items in which the company has

manufacturing experience.

9. Criteria for buy

The following are the criteria for buy:

15. Requires high investments on facilities which are already available at suppliers plant.

16. The company does not have facilities to make it and

there are more profitable opportunities for investing

company’s capital.

17. Existing facilities of the comp ny c n be used more economically to make other p r s.

18. The skill of personnel employed by the company is not

readily adaptable to make the p rt.

19. Patent or other legal b rriers prevent the company for making the

part.

20. Demand for the part is either temporary or seasonal.

2.3 APPROACHES FOR MAKE OR BUY DECISION

Types of analysis followed in make or buy decision are as follows:

1. Simple cost analysis

2. Economic analysis

3 Break-even analysis

1. Simple Cost Analysis

The concept is illustrated using an example problem.

EXAMPLE

A company has extra capacity that can be used to produce a sophisticated

fixture which it has been buying for Rs. 900 each. If the company makes the

fixtures, it will incur materials cost of Rs. 300 per unit, labour costs of Rs. 250

per unit, and variable overhead costs of Rs. 100 per unit. The annual fixed cost

associated with the unused capacity is Rs. 10,00,000. Demand over the next year

is estimated at 5,000 units. Would it be profitable for the company to make the

fixtures?

Solution

We assume that the unused capacity has alternative use.

Cost to make Variable cost/unit = Material + labour + overheads

= Rs. 300 + Rs. 250 + Rs. 1 0 0

= Rs. 650

Total variable cost = (5,000 units) (Rs. 650/unit)

= Rs. 32,50,000

Add fixed cost associated

with unused capacity + Rs. 10,00,000

Total cost = Rs. 42,50,000

Cost to buy

Purchase cost = (5,000 units) (Rs. 900/unit)

Add fixed cost associated = Rs. 45,00,000

with unused capacity + Rs. 10,00,000

Total cost = Rs. 55,00,000

The cost of making fixtures is less than the cost of buying fixtures from

outside. Therefore, the organization should make the fixtures.

2. Economic Analysis

The following inventory models are considered to illustrate this concept:

Purchase model

Manufacturing model

The formulae for EOQ and total cost (TC) for each model are

given in the following table:

Purchase model com

Manufacturing model

2C o D

2CD

Q1 =

Cc

Q2 = --------

Cc (1 − r /k)

TC = D P + DC

o + Q1 × Cc TC = D P + DC

Q2 Q1 2

+ Cc (k – r) Q2

2 * k

where

D = demand/year

P = purchase price/unit

Cc = carrying cost/unit/year

Co = ordering cost/order or set-up cost/set-up

k = production rate (No. of units/year)

r = demand/year

Q1 = econom c order size

Q2 = econom c production size

TC = total cost per year

EXAMPLE

An item has a yearly demand of 2,000 units. The different costs in

respect of make and buy are as follows. Determine the best option.

Buy Make

Item cost/unit Rs. 8.00 Rs. 5.00

Procurement cost/order Rs. 120.00

Set-up cost/set-up Rs. 60.00

Annual carrying cost/

item/year Rs. 1.60 Rs. 1.00

Production rate/year 8,000 units

Solution

Buy option

CiviQ2ldatas2r

D = 2,000 units/year

Co = Rs. 120/order

Cc = Rs. 1.60/unit/year

Q1 = 2Co D

= 2 × 2,000 × 120

Cc

1.60

= 548 units (approx.)

DC o Q1C c

TC = DP +

Q1 +

2

= 2,000

8 + 2,000 × 120 +

548 × 1.60

548 2

= Rs. 16,876.36

Make option

Co = Rs. 60/set-up r = 2,000 units/year

Cc = Re 1/unit/year

k = 8,000 units/year

-

o

Cc[1 − (r / k)]

=

2 × 60 × 2,000 = 566 units (approx.)

1.0(1 − 2,000/8,000)

C = DP + D × C

o + Cc (k – r) Q2

2 × k

Q2

2,000 × 60 566 = 2,000 5.00 + + 1.0 (8,000 – 2,000)

566 2 × 8,000

= Rs. 10,424.26

Result: The cost of making is less than the cost of buying. Therefore, the firm should go in for the making option.

2.5 Aims The aims of value engineering are as follows:

✓ Simplify the product.

✓ Use (new) cheaper and better materials.

✓ Modify and improve product design.

✓ Use efficient processes.

✓ Reduce the product cost.

✓ Increase the utility of the product by economical means.

✓ Save money or increase the profits.

2.6 Value Engineering Procedure

The basic steps of value engineering are as follows:

(a) Blast (i) Identify the product. (ii) Collect relev nt information. (iii) Define different functions.

(b) Create (iv) Different lternatives. (v) Critic lly ev lu te the alternatives.

(c) Refine (vi) Develop the best alternative.

(vii) Imp ement the alternative.

Step 1: Identify the product. F rst, dent fy the component for study. In future, any

design change should add value and it should not make the product as obsolete one. Value

engineering can be applied to a product as a whole or to sub-units.

Step 2: Collect relevant information. Information relevant to the following must be collected:

Technical specifications with drawings

✓ Production processes, machine layout and instruction sheet

✓ Time study details and manufacturing capacity

✓ Complete cost data andmarketing details

✓ Latest development in related products

Step 3: Define different functions. Identify and define the primary, secondary and tertiary functions of

the product or parts of interest. Also, specify the value content of each function and identify the high cost

areas.

Step 4: Different alternatives. Knowing the functions of each component part and its

manufacturing details, generate the ideas and create different alternatives so as to increase the value of

the product.Value engineering should be done after a brain storming session. All feasible or non-feasible

suggestions are recorded without any criticism; rather, persons are encouraged to express their views

freely.

2.7 Interest Formulas Interest rate can be classified into simple interest rate and compound

interest rate.

✓ In simple interest, the interest is calculated, based on the initial deposit for every interest period. In this case, calculation of interest on interest is not applicable.

✓ In compound interest, the interest for the current period is computed based on the amount (principal plus interest up to the end of the previous period) at the beginning

of the current period.

The notations which are used in various interest formulae are as follows:

P = principal amount

n = No. of interest periods

5. = interest rate (It may be compounded monthly, quarterly, semiannually or annually)

F = future amount at the end of year n

A = equal amount deposited at the end of every interest period

G = uniform amount which will be added/subtracted period after period to/from the amount of depo it A1 at the end of period 1

2.8 Time Value of Money - . It represents the growth of capital per unit period. The period

may be a month, a quarter, semiannual or year.

An interest rate 15% compounded annually means that for every hundred

rupees invested now, an amount of Rs. 15 will be added to the account at the end

of the first year. So, the total amount at the end of the first year will be Rs. 115.

At the end of the second year, again 15% of Rs. 115, i.e. Rs. 17.25 will be added to the account.

Hence the total amount at the end of the second year will be Rs. 132.25. The process will continue thus till the specified number of years.

If an investor invests a sum of Rs. 100 in a fixed deposit for five years with an interest rate of 15% compounded annually, the accumulated amount at the end of

every year ill be as shown in Table

Compound Amounts

(amount of deposit = Rs. 100.00)

Year end Interest Compound amount

(Rs.) (Rs.)

0 100.00 1 15.00 115.00

2 17.25 132.25

3 19.84 152.09

4 22.81 174.90

5 26.24 201.14

The formula to find the future worth in the third column is

F = P (1 + i)n

where P = principal amount invested at time 0, F = future amount, i = interest rate compounded annually, n = period of deposit.

The maturity value at the end of the fifth year is Rs. 201.14. This means that

the amount Rs. 201.14 at the end of the fifth year is equivalent to Rs. 100.00 at

time 0 (i.e. at present). This is diagrammatically shown in Fig. 3.1. This

explanation assumes that the inflation is at zero percentage.

2

2.10 Single-Payment Present Worth Amount - Here, the objective is to find the present worth

amount (P) of a single future sum (F) which will be received after n periods at an interest rate

of i compounded at the end of every interest period. F

. .

0 1 23 4. . n

P i % Cash flow diagram of single-payment present worth amount.

The formula to obtain the present worth is

P = F = F(P/F, i, n)

(1 + i)n

Where (P/F, i, n) is termed as single-payment present worth factor.

2.11 Equal-Payment Series Sinking Fund

In this type of investment mode, the objective is to find the equivalent amount

(A) that should be deposited at the end of every interest period for n interest

periods to realize a future sum (F) at the end of the nth interest period at an

interest rate of i.

i % n

0 1 2 3 4. . .

.

A A A A i %

A = equal amount to be deposited at the end of ea h interest period n = No. of interest periods

A = annual equivalentCivildatapayments

i = rate of interest

F = single future amount at the end of the nth period

The formula to get F is

A = F = F(A/F, i, n)

(1 + )n − 1

Where

(A/F, i, n) is called as equal-payment series sinking fund factor.

2.12 Equal-Payment Series Present Worth Amount The objective of this mode of investment is

find the present worth of an equal payment made at the end of every interest period for

interest periods at an interest rate of i compounded at the end of every interest period.

The corresponding cash flow diagram is shown in Fig. 3.8.

Here, P = present worth

i = interest rate n = No. of interest periods

The formula to compute P is

to n

2 = A (1 +

i)n

−

1

= A(P/A, i, n) i(1 +

i)n

Where (P/A, i, n) is called equal-payment series present worth factor

2.13 Equal-Payment Series capital recovery The objective of this mode of investment is to

find the annual equivalent amount (A) which is to be recovered at the end of every interest

period for n interest periods for a loan (P) which is sanctioned now at an interest rate of i

compounded at the end of every interest period

P

6. %

0 1 2 3 4 . .

n

. .

A

A

A

A

A Cash flow diagram of equal-payment series capital recovery a ount.

P = present worth (loan amount) A = annual equivalent payment (recovery amount) i = interest rate n = No. of interest periods

Civildatas

The formula to compute P is as follow :

(1 + i)n

A = P

= P(A/P, i, n)

(1 + i)n − 1

Where,

(A/P, i, n) is called equal-p yment series capital recovery factor.

2.14 Uniform Gradient series annual equivalent The objective of this mode of investment is

to find the annual equivalent amount of a series with an amount A1 at the end of the first

year and with an equal increment (G) at the end of each of the following n – 1 years with an

interest rate i compounded annually.

The correspond ng cash flow diagram is shown in Fig

0 1 2 3 4 . . 10 . .

A1

A1+G

A1+2G

A1+3G

A1+ (n – 1)G

Cash flow diagram of uniform gradient series annual equivalent amount.

The formula to compute A under this situation is

(1 + i)n

− in − 1 A = A1 + G

i(1 + i)n

− i = A1 + G (A/G, i, n)

here Where

(A/G, i, n) is called uniform gradient series factor.

2.15 Effective Interest rate Let i be the nominal interest rate compounded annually. But, in

practice, the compounding may occur less than a year. For example, compounding may

be monthly, quarterly, or semi-annually. Compounding monthly means that the interest is

computed at the end of every month. There are 12 interest periods in a year if the interest

is compounded monthly. Under such situations, the formula to compute the effective

interest rate, which is compounded annually, is Effective interest rate, R = 1 +

i/C C

− 1

where,

i = the nominal interest rate C = the number of interest periods in a year.

UNIT III CASH FLOW

3.1 Introduction

In this method of comparison, the cash flows of each alternative will be reduced to time zero by

assuming an interest rate i. Then, depending on the type of decision, the best alternative will be

selected by comparing the present worth amounts of the alternatives.

The sign of various amounts at different points in time in a cash flow diagram is to be decided based on the type of the decision problem.

In a cost dominated cash flow diagram, the costs (outflows) will be assigned with positive sign

and the profit, revenue, salvage value (all inflows), etc. will be assigned with negative sign.

In a revenue/profit-dominated cash flow diagram, the profit, revenue, salvage value (all inflows to an organization) will be assigned with positive sign The costs (outflows) will be assigned with

negative sign.

In case the decision is to select the alternative with the minimum cost, then the alternative with the least present worth amount will be selec ed. On the other hand, if the decision is to select the

alternative with the maximum profi , hen he alternative with the maximum present worth will be selected.

3.2 BASES FOR COMPARISON OF ALTERNATIVES

In most of the practical dec s on environments, executives will be forced to select the best alternative from a set of compet ng alternatives.

Let us assume that an organ zation has a huge sum of money for potential investment and there

are three different projects whose initial outlay and annual revenues during their lives are known. The executive has to select the best alternative among these three competing projects.

There are several bases for comparing the worthiness of the projects. These bases are:

= Present worth method = Future orth method = Annual equivalent method = Rate of return method

3.2.1 PRESENT WORTH METHOD

In this method of comparison, the cash flows of each alternative will be reduced to time zero by assuming an interest rate i.

Then, depending on the type of decision, the best alternative will be selected by comparing the present worth amounts of the alternatives.

In a cost dominated cash flow diagram, the costs (outflows) will be assigned with positive sign and the profit, revenue, salvage value (all inflows), etc. will be assigned with negative sign.

In a revenue/profit-dominated cash flow diagram, the profit, revenue,

salvage value (all inflows to an organization) will be assigned with positive sign. The costs (outflows) will be assigned with negative sign.

3.1.1.1 Revenue-Dominated Cash Flow Diagram

A generalized revenue-dominated cash flow diagram to demonstrate the present worth method of comparison is presented in Fig.

S

Rj

R1 R2 R3 Rn

.

0 1 2 3 j n

To find the present worth of the bove cash flow diagram for a given interest rate, the formula is

PW(i) = – P + R1[1/(1 + i)1

] + R2[1/(1 + i)2

] + ...

+ Rj [1/(1 + i) j] + Rn[1/(1 + i)

n] + S[1/(1 + i)

n]

3.1.1.2 Cost-Dominated Cash Flow D agram

A generalized cost-dominated cash flow diagram to demonstrate the present worth method of comparison s presented in Fig.

S

0 1 2 . j

C1 C2 . Cj Cn

P

To compute the present worth amount of the above cash flow diagram for a given interest rate i, e have the formula

PW(i) = P + C1[1/(1 + i)1

] + C2[1/(1 + i)2

] + ... + Cj[1/(1 + i) j]

+ Cn[1/(1 + i)n

] – S[1/(1 + i)n

]

SCE 33 Department of Mechanical Engineering

Visit : www.Civildatas.com Visit : www.Civildatas.com

EXAMPLE

Alpha Industry is planning to expand its production operation. It has identified three different technologies for

meeting the goal. The initial outlay and annual revenues with respect to each of the technologies are summarized in

Table 1. Suggest the best technology which is to be implemented based on the present worth method of comparison

assuming 20% interest rate, compounded annually.

Table 1

Initial outlay Annual revenue Life

(Rs.) (Rs.) (years) Technology 1 12,00,000 4,00,000 10 Technology 2 20,00,000 6,00,000 10 Technology 3 18,00,000 5,00,000 10

Solution In all the technologies, the initial outlay is assigned a negative sign and the annual

revenues are assigned a positive sign

TECHNOLOGY 1

Initial outlay, P = Rs. 12,00,000 Annual revenue, A = Rs. 4,00,000 Interest rate, i = 20%, compounded nnu lly Life of this technology, n = 10 years

The cash flow diagram of this technology is as shown in Fig. 4.3.

4,00,000 4,00,000 4,00,000 . . 4,00,000

0 1 2 3 . j 10

=20%

12,00,000

F g. Cash flow diagram for technology 1.

The present worth expression for this technology is

PW(20%)1 = –12,00,000 + 4,00,000 (P/A, 20%, 10)

= –12,00,000 + 4,00,000 (4.1925)

Civildatas = –12,00,000 + 16,77,000

= Rs. 4,77,000

TECHNOLOGY 2

Initial outlay, P = Rs. 20,00,000 Annual revenue, A = Rs. 6,00,000 Interest rate, i = 20%, compounded annually Life of this technology, n = 10 years

The cash flow diagram of this technology is shown in Fig. 4.4.

6,00,000 6,00,000 6,00,000

.

.

0 1

10

2 i = 20%

20,00,000

Fig. Cash flow diagram for technology 2.

The present worth expression for this technology is

PW(20%)2 = – 20,00,000 + 6,00,000 (P/A, 20%, 10)

= – 20,00,000 + 6,00,000 (4.1925)

= – 20,00,000 + 25,15,500

= Rs. 5,15,500

TECHNOLOGY 3

Initial outlay, P = Rs. 18,00,000 Annual revenue, A = Rs. 5,00,000 Interest rate, i = 20%, compoun ed annually Life of this technology, n = 10 years

The cash flow diagram of th s technology is shown in Fig. 4.5.

18,00,000CivildatasFig.Cashflowdiagramfortechnology 3.

5,00,000 5,00,000 5,00,000

.

.

0 1

10 2

=20%

The present orth expression for this technology is

PW(20%)3 = –18,00,000 + 5,00,000 (P/A, 20%, 10)

15. –18,00,000 + 5,00,000 (4.1925)

a. –18,00,000 + 20,96,250

b. Rs. 2,96,250

From the above calculations, it is clear that the present worth of technology 2 is the

highest among all the technologies. Therefore, technology 2 is suggested for implementation to

expand the production.

3.1.2 FUTURE WORTH METHOD

In the future worth method of comparison of alternatives, the future worth of various alternatives will be computed.

Then, the alternative with the maximum future worth of net revenue or with the minimum future worth of net cost will be selected as the best alternative for implementation.

3.1.2.1 Revenue-Dominated Cash Flow Diagram

A generalized revenue-dominated cash flow diagram to demonstrate the future worth method of comparison is presented in Fig.

S

R1 R2 R3 Rj Rn

0 1 2 3 . j n

P

Fig Revenue-dominated c sh flow diagram.

In Fig. P represents an initial investment, Rj the net-revenue at the end of the jth year, and S the salvage value at the end of the nth year.

The formula for the future worth of the above cash flow diagram for a given interest rate, i is

FW(i) = –P(1 + )n

+ R1(1 + i)n–1

+ R2(1 + i)n–2

+ ...

+ R j(1 + i)n–j

+ ... + Rn + S

In the above formula, the expenditure is assigned with negative sign and the revenues are assigned ith positive sign.

3.1.2.2 Cost-Dominated Cash Flow Diagram

A generalized cost-dominated cash flow diagram to demonstrate the future worth method of comparison is given in Fig.

S

0 1 2 j

C1 C2 Cj Cn

P

Fig. Cost-dominated cash flow diagram.

In Fig. 5.2, P represents an initial investment, Cj the net c st of operation and

maintenance at the end of the j th year, and S the salvage value at the end of the nth year.

The formula for the future worth of the above cash flow diagram for a given interest rate, i is

FW(i) = P(1 + i)n

+ C1(1 + i )n–1

+ C2(1 +

i)n–2

+ ...

+ Cj(1 + i)n–j

+ ... + Cn – S

EXAMPLE

Consider the following two mutually exclusive alternatives:

End of year

Alternative 0 1 2 3 4

A (Rs.) – 50,00,000 20,00,000 20,00,000 20,00,000 20,00,000

B (Rs.) – 45,00,000 18,00,000 18,00,000 18,00,000 18,00,000

At i = 18%, select the best alternative based n future worth method of

comparison.

Solution Alternative A Civildatas

Initial investment, P = Rs. 50,00,000

Annual equivalent revenue, A = Rs. 20,00,000

Interest rate, i = 18%, compounded nnu lly

Life of alternative A = 4 years

The cash flow diagram of altern tive A is shown in Fig.

20,00,000 20,00,000 20,00,000 20,00,000

0 1 2 3 4

=18%

50,00,000

Fig. Cash flow diagram for alternative A.

The future worth amount of alternative B is computed as

FWA(18%) = –50,00,000(F/P, 18%, 4) + 20,00,000(F/A, 18%, 4)

= –50,00,000(1.939) + 20,00,000(5.215)

= Rs. 7,35,000

Alternative B

Initial investment, P = Rs. 45,00,000 Annual equivalent revenue, A = Rs. 18,00,000 Interest rate, i = 18%, compounded annually Life of alternative B = 4 years

The cash flow diagram of alternative B is illustrated in Fig..

18,00,000 18,00,000 18,00,000 18,00,000

0 1 2 3 4

i =18%

45,00,000

Fig. Cash flow diagram for alternative B.

The future worth amount of alternative B is computed as

FWB(18%) = – 45,00,000(F/P, 18%, 4) + 18,00,000 (F/A, 18%, 4)

= – 45,00,000(1.939) + 18,00,000(5.215)

= Rs. 6,61,500

3.1.3 ANNUAL EQUIVALENT METHOD

✓ In the annual equivalent method of comparison, first the annual equivalent cost or the revenue of each alternative will be computed.

✓ Then the alternative with the maximum annual equivalent revenue in the case of revenue-

based comparison or with the minimum annual equivalent cost in the case of cost- based comparison will be selected as the best alternative.

3.1.3.1 Revenue-Dominated Cash Flow Diagram

A generalized revenue-dominated cash flow diagram to de equivalent method of comparison is presented in Fig.

S

R1 R2 R3 Rj Rn

.

0 1 2 3 j

P

Fig. Revenue-domin ted c sh flow diagram.

In Fig. P represents an initial investment, Rj the net revenue at the end of the j th year, and S the salvage va ue at the end of the nth year.

The first step is to f nd the net present worth of the cash flow diagram using the following expression for a given interest rate, i:

PW(i) = –P + R1/(1 + i)1

+ R2/(1 + i)2

+ ...

+ Rj/(1 + i) j + ... + Rn/(1 + i)

n + S/(1 + i)

n

In the above formula, the expenditure is assigned with a negative sign and the revenues are assigned with a positive sign.

3.1.3.2 Cost-Dominated Cash Flow Diagram

A generalized cost-dominated cash flow diagram to demonstrate the annual equivalent method of comparison is illustrated in Fig.

S

0 1 2 .

C1 C2 Cj Cn

P

In Fig, P represents an initial investment, Cj the net ost of operation and maintenance at the end of the jth year, and S the salvage value at the end of the nth year.

The first step is to find the net present worth of the ca h flow diagram using the following relation for a given interest rate, i.

PW(i) = P + C1/(1 + i)1

+ C2/(1 + i)2

+ ...

+ Cj/(1 + i) j + ... + Cn/(1 + i)

n – S/(1 + i)

n

EXAMPLE

A company provides a car to its chief executive. The owner of the company is

concerned about the increasing cost of petrol. The cost per litre of petrol for the first

year of operation is Rs. 21. He feels that the cost of petrol will be increasing by Re.1

every year. His experience with his company car indicates that it averages 9 km per litre

of petrol. The executive expects to drive an average of 20,000 km each year for the next

four years. What is the annual equivalent cost of fuel over this period of time?. If he is

offered similar service with the same quality on rental basis at Rs. 60,000 per year,

should the owner continue to provide company car for his executive or alternatively

provide a rental car to his executive? Assu e i = 18%. If the rental car is preferred, then

the company car will find so e other use within the company.

Solution

Average number of km run/year = 20,000 km

Number of km/litre of petrol = 9 km

Therefore,

Petrol consumption/year = 20,000/9 = 2222.2 litre Cost/litre

of petrol for the 1st ye r = Rs. 21

Cost/litre of petrol for the 2nd ye r = Rs. 21.00 + Re. 1.00 = Rs. 22.00

Cost/litre of petrol for the 3rd year = Rs. 22.00 + Re. 1.00 = Rs. 23.00 Cost/litre of petrol for the 4th year = Rs. 23.00 + Re. 1.00 = Rs. 24.00 Fuel expenditure for 1st year = 2222.2 21 = Rs. 46,666.20 Fuel expend ture for 2nd year = 2222.2 22 = Rs. 48,888.40 Fuel expend ture for 3rd year = 2222.2 23 = Rs. 51,110.60 Fuel expenditure for 4th year = 2222.2 24 = Rs. 53,332.80

The annual equal increment of the above expenditures is Rs. 2,222.20 (G). The

cash flow diagram for this situation is depicted in Fig.

0 1 2 3 4

A1

A1 + G A1 +2G

A1 +3G

Fig. Uniform gradient series cash flow diagram.

In Fig., A1 = Rs. 46,666.20 and G = Rs. 2,222.20

A = A1 + G(A/G, 18%, 4) = 46,666.20 + 2222.2(1.2947)

= Rs. 49,543.28

The proposal of using the company car by spending for petrol by the company

will cost an annual equivalent amount of Rs. 49,543.28 for four years. This amount

is less than the annual rental value of Rs. 60,000. Therefore, the company should

continue to provide its own car to its executive.

3.1.4 RATE OF RETURN METHOD

✓ The rate of return of a cash flow pattern is the interest rate at which the present worth of that cash flow pattern reduces to zero.

✓ In this method of comparison, the rate of return for each alternative is computed. Then the alternative which has the highest rate of return is selected as the best alternative.

✓ A generalized cash flow diagram to demonstrate the rate of return method of comparison is presented in Fig

S

R1 R2 R3 . Rj Rn

0 1 2 3 . j n

P

Fig. Generalized cash flow diagram

In the above cash flow diagram, P represents n initi l investment, Rj the net revenue at the end of the jth year, and S the salvage value at the end of the nth year.

The first step is to find the net present worth of the cash flow diagram using the following expression at a given interest rate, i.

PW(i) = – P + R1/(1 + i)1

+ R2/(1 + i)2

+ ...

+ Rj/(1 + i) j + ... + Rn/(1 + i)

n + S/(1 + i)

n

UNIT IV REPLACEMENT AND MAINTENANCE ANALYSIS

4.1 Introduction

Organizations providing goods/services use several facilities like equipment and machinery which are directly required in their operations. In addition to these facilities,

there are several other items which are necessary to facilitate the functioning of

organizations.

All such facilities should be continuously monitored for their efficient functioning; otherwise, the quality of service will be poor. Besides the quality of service of the

facilities, the cost of their operation and maintenance would increase with the passage of time.

Hence, it is an absolute necessity to maintain the equipment in good operating conditions with economical cost. Thus, we need an integrated approach to minimize the cost of maintenance. In certain cases, the equipment will be obsolete over a period of time.

If a firm wants to be in the same business competitively, it has to take decision on whether to replace the old equipment or to re in it by t king the cost of maintenance and operation into account.

There are two basic reasons for considering the replacement of an equipment physical impairment of the various parts or obsolescence of the equipment.

Physical impairment refers on y to changes in the physical condition of the machine

itself. This would lead to a dec ne n the value of the service rendered, increased operating cost, increased maintenance cost or a combination of these.

Obsolescence is due to mpro ement of the tools of production, mainly improvement in technology.

So, it would be uneconomical to continue production with the same machine under any of the above situations. Hence, the machines are to be periodically replaced.

Sometimes, the capacity of existing facilities may be inadequate to meet the current demand. Under such situation, the following alternatives will be considered.

✓ Replacement of the existing equipment with a new one.

✓ Augmenting the existing one with an additional equipment.

4.2 Types of Maintenance

Maintenance activity can be classified into two types:

✓ Preventive maintenance and

✓ Breakdown maintenance.

Preventive maintenance (PM) is the periodical inspection and service activities which are aimed to detect potential failures and perform minor adjustments or repairs which will prevent major operating problems in future.

Breakdown maintenance is the repair which is generally done after the equipment has attained down state. It is often of an emergency nature which will have associated penalty in terms of expediting cost of maintenance and down time cost f equipment.

Preventive maintenance will reduce such cost up to a point Beyond that point, the cost of preventive maintenance will be more when compared to the breakdown maintenance cost.

The total cost, which is the sum of the preven ive m intenance cost and the breakdown maintenance cost, will go on decreasing with an increase in the level of maintenance up to a point.

Beyond that point, the total cost will start increasing. The level of maintenance

corresponding to the minimum total cost is the optimal level of maintenance. The concepts are demontrated in F g.

4.3 Types of Replacement Problem

Replacement study can be classified into two categories:

(a) Replacement of assets that deteriorate with time (Replacement due to gradual failure, or wear and tear of the components of the machines).

This can be further classified into the following types:

(i) Determination of economic life of an asset.

(ii) Replacement of an existing asset with a new asset.

(b) Simple probabilistic model for assets whi h fail ompletely (replacement due to sudden failure).

4.4 Determination of Economic Life of an Asset

Any asset will have the following cost components:

Capital recovery cost (average first cost), computed from the first cost (purchase price) of the machine.

Average operating and ma ntenance cost (O & M cost)

Total cost which is the sum of capital recovery cost (average first cost) and average maintenance cost.

EXAMPLE

A firm is considering replacement of an equipment, whose first cost is Rs. 4,000 and

the scrap value is negligible at the end of any year. Based on experience, it was found that the maintenance cost is zero during the first year and it increases by Rs. 200 every

year thereafter.

(a) When should the equipment be replaced if i = 0%?

(b) When should the equipment be replaced if i = 12%?

(a) When i = 0%. In this problem

(i) First cost = Rs. 4,000

(ii) Maintenance cost is Rs. 0 during the first year and it increases by Rs. 200 every year thereafter

This is summarized in column B of Table

Column C summarizes the summation of maintenance costs for each replacement

period. The value corresponding to any end of year in this column represents the total

maintenance cost of using the equipment till the end of that particular year.

Average total cost = First cost (FC) + Summation of maintenance cost

Replacement period

year. com = n n

= Average first cost

+ Average aintenance

he given period cost f r the given for

peri d

Column F = Column E + Column D

The value corresponding to any end of year (n) in Column F represents the

average total cost of using the equipment till the end of that particular

For this problem, the average tot l cost decreases till the end of year 6 and then it increases. Therefore, he optimal replacement period is six years, i.e. economic life of the equipmen is six years.

(b) When interest rate, i = 12%. When the interest rate is more than 0%, the

steps to be taken for getting the economic life are summarized with reference to

Table

Table Calculat ons to Determine Economic Life (First cost = Rs. 4,000,

Interest = 12%)

End of Maintenance P/F, 12%, Present worth as of Summation of Present worth of A/P, 12%, Annual

year cost at end of n beginning of year 1 present worth cumulative n equivalent

year of maintenance of maintenance maintenance cost total

(n) costs costs through year & first cost cost through

given year given

(B C) D E + Rs. 4,000 F G

A B (Rs) C D (Rs.) E (Rs.) F (Rs.) G H (Rs.)

1 0 0.8929 0.00 0.00 4,000.00 1.1200 4,480.00

2 200 0.7972 159.44 159.44 4,159.44 0.5917 2,461.14

3 400 0.7118 284.72 444.16 4,444.16 0.4163 1,850.10

4 600 0.6355 381.30 825.46 4,825.46 0.3292 1,588.54

5 800 0.5674 453.92 1,279.38 5,279.38 0.2774 1,464.50

6 1,000 0.5066 506.60 1,785.98 5,785.98 0.2432 1,407.15

7 1,200 0.4524 542.88 2,328.86 6,328.86 0.2191 1,386.65*

8 1,400 0.4039 565.46 2,894.32 6,894.32 0.2013 1,387.83

9 1,600 0.3606 576.96 3,471.28 7,471.28 0.1877 1,402.36

10 1,800 0.3220 579.60 4,050.88 8,050.88 0.1770 1,425.00

*Economic life of the machine = 7 years

The steps are summarized now:

1. Discount the maintenance costs to the beginning of year 1.

Column D = Column B

1 (1 + i)

n

= Column B (P/F, i, n) = Column B Column C.

2. Find the summation of present worth of maintenance costs

through the year given (Column E = Column D).

3. Find Column F by adding the first cost of Rs. 4,000 to Column E.

4. Find the annual equivalent total cost through the years given.

ColumnCivildatasH=ColumnF(1+i)n−

(1 + i)n 1

= Column F (A/P, 12%, n) = Column F Column G

5. Identify the end of ye r for which the annual equivalent total

cost is minimum.

For this problem, the annual equivalent total cost is minimum at the end of year 7.

Therefore, the economic ife of the equipment is seven years.

REPLACEMENT OF EXIST ING ASSET WITH A NEW ASSET

In this section, the c oncept of comparison of replacement of an existing asset with a new asset is presented. In this analysis, the annual equivalent cost of each alternative

should be co mputed first. Then the alternative which has the least cost should be selected as the best alternative.

Before discu ssing details, some preliminary concepts whic h are essential for this type of

replacem ent analysis are presented.

Capital Recovery with Retur n

Consider the follow ing data of a machine.

Let

P = purchase priice of the machine,

F = salvage valu e of the machine t the end of machine life,

n = life of the m achine in years, nd

i = interest rate,, compounded annually

The corresponding cash flow diagram is shown in Fig

The equation for the annual equivalent amount for the above cash flow diagram is

AE(i) = (P – F ) (A/P, i, n) + F i

This equation represents the capital recovery with return.

Concept of Challenger and Defender

o If an existing equipment is considered for replacement with a new equipment, then the existing equipment is known as the defender and the new equipment is known as challenger.

o Assume that an equipment has been purchased about three years back for Rs. 5,00,000 and it is considered for replacement with a new equipment. The supplier of the new equipment will take the old one for some money, say, Rs. 3,00,000.

o This should be treated as the present value of the existing equipment and it should be considered for all further economic analysis.

o The purchase value of the existing equipment before three years is now known as sunk cost, and it should not be onsidered for further analysis.

EXAMPLE

Two years ago, a machine was purch sed t a cost of Rs. 2,00,000 to be useful for eight years. Its salvage value a he end of its life is Rs. 25,000. The annual maintenance

cost is Rs. 25,000. The market value of the present m chine is Rs. 1,20,000. Now, a new machine to

cater to the need of the present machine is available at Rs. 1,50,000 to be useful for six

years. Its annual maintenance cost is Rs. 14,000. The salvage value of the new machine is Rs. 20,000.

Using an interest rate of 12%, find whether it is worth replacing the present machine with the new mach ne.

Solution

Alternative 1—

Present machine

Purchase price = Rs. 2,00,000

Present value (P) = Rs. 1,20,000

Salvage value (F) = Rs. 25,000

Annual maintenance cost (A) = Rs. 25,000

Remaining life = 6 years

Interest rate = 12%

The cash flow diagram of the present machine is illustrated in Fig.

25,000

–2 –1 0 1 2 3 4 5 6

25,000 25,000 25,000 25,000 25,000 25,000

1,20,000 2,00,0002,20,000

Fig. Cash flow diagram for alternative 1.

annual maintenance cost for the preceding periods are not shown in

this figure. The annual equivalent cost is computed as

AE(12%) = (P – F)(A/P, 12%, 6) + F i + A

= (1,20,000 – 25,000)(0.2432) + 25,000 0.12 + 25,000

= Rs. 51,104

Alternative 2 —

New machine

Purchase price (P) = Rs. 1,50,000 Salvage value (F) = Rs. 20,000 Annual maintenance cost (A) = Rs. 14,000 Life = 6 years Interest rate = 12%

The cash flow diagram of the new machine is depicted in Fig.

20,000

0 1 2 3 4 5 6

14,000 14,000 14,000 14,000 14,000 14,000

1,50,000

Fig. Cash flow diagram for alternative 2.

The formula for the annual equivalent cost is

AE(12%) = (P – F)(A/P, 12%, 6) + F i + A

= (1,50,000 – 20,000)(0.2432) + 20,000 0.12 + 14,000

= Rs. 48,016

Since the annual equivalent cost of the new machine is less than that of the

present machine, it is suggested that the present machine be replaced with the new

machine.

Simple Probabilistic Model For Items Which Fail Completely

Electronic items like transistors, re i tor , tubelights, bulbs, etc. could fail all of a

sudden, instead of gradual deterioration. The failure of the item may result in

complete breakdown of the system. The system may contain a collection of such

items or just one i em, s y tubelight.

Therefore, we use some repl cement policy for such items which would avoid

the possibility of a complete bre kdown.

The following are the rep acement policies which are applicable for this

situation.

(i) Individual replacement policy. Under this policy, an item is replaced

immediately after ts failure.

(ii) Group replacement policy. Under this policy, the following decision is made:

At what equal intervals are all the items to be replaced simultaneously with a

provision to replace the items individually which fail during a fixed group

replacement period?

There is a trade-off between the individual replacement policy and the group

replacement policy. Hence, for a given problem, each of the replacement policies is

evaluated and the most economical policy is selected for implementation. This is

explained with two numerical problems.

EXAMPLE

The failure rates of transistors in a computer are summarized in

Table .

Table Failure Rates of Transistors in Computers

End of week 1 2 3 4 5 6 7

Probability of 0.07 0.18 0.30 0.48 0.69 0.89 1.00

failure to date

The cost of replacing an individual failed transistor is Rs. 9. If all

the transistors are replaced simultaneously, it w uld c st Rs. 3.00 per

transistor. Any one of the following tw pti ns can be followed to replace

the transistors:

(a) Replace the transistors individually when they fail (individual

replacement policy).

(b) Replace all the transistors simult neously at fixed intervals and

replace the individual r nsis ors s they fail in service during the

fixed interval (group replacement policy).

Find out the optimal repl cement policy, i.e. individual replacement

policy or group repl cement policy. If group replacement policy is

optimal, then find at what equal intervals should all the transistors be

replaced.

Solution

Assume that there are 100 transistors in use.

Let,

pi be the probability that a transistor which was new when placed in

position for use, fails during the ith week of its life. Hence, p1 = 0.07, p2 = 0.11, p3 = 0.12, p4 = 0.18,

p5 = 0.21, p6 = 0.20, p7 = 0.11

Since the sum of pis is equal to 1 at the end of the 7th week, the

transistors are sure to fail during the seventh week.

Assume that (a) transistors that fail during a week are replaced just before the

end of the week, and (b) the actual percentage of failures during a week for a sub-group

of transistors with the same age is same as the

expected percentage of failures during the week for that sub-group of

transistors.

Let Ni = the number of transistors replaced at the end of the ith week

N0 = number of transistors replaced at the end of the week 0 (or at the

beginning of the first week).

= 100

N1 = number of transistors replaced at the end of the 1st week = N0 p1 = 100 0.07 = 7 N2 = number of transistors replaced at the end of the 2nd week = N0 p2 + N1 p1

= 100 0.11 + 7 0.07 = 12 N3 = N0 p3 + N1 p2 + N2 p1

= 100 0.12 + 7 0.11 + 12 0.07 = 14Civildatas

N4 = N0 p4 + N1 p3 + N2 p2 + N3 p1 = 100 0.18 + 7 0.12 + 12 0.11 + 14 0.07 = 21

N5 = N0 p5 + N1 p4 + N2 p3 + N3 p2 + N4 p1 = 100 0.21 + 7 0.18 + 12 0.12 + 14 0.11 + 21 0.07 = 27

= 100 0.2 + 7 0.21 + 12 0.18 + 14 0.12 + 21 0.11 + 27 = 30

N7 = N0 p7 + N1 p6 + N2 p5 + N3 p4 + N4 p3 + N5 + N6 p1

= 100 0.11 + 7 0.2 + 12 0.21 + 14 0.18 + 21 0.12 + 27 0.11 + 30 0.07 = 25

UNIT V- DEPRECIATION

5.1 Introduction

✓ Any equipment which is purchased today will not work for ever. This may be due to wear and tear of the equipment or obsolescence of technology.

✓ Hence, it is to be replaced at the proper time for continuance of any business.

✓ The replacement of the equipment at the end of its life involves oney. This must be internally generated from the earnings of the equip ent.

The recovery of money from the earnings of an equipment for its replacement

purpose is called depreciation fund since we make an assumption that the value of the equipment decreases with the passage of time.

Thus the word “depreciation” means decrea e in value of any physical asset

with the passage of time.

5.2 Methods of Depreciation

There are several methods of accounting depreciation fund. These are as follows:

1. Straight line method of epreciation

2. Declining balance method of depreciation

3. Sum of the years—digits method of depreciation

4. Sinking-fund method of depreciation

5. Service output method of depreciation

1. Straight Line Method of Depreciation

In this method of depreciation, a fixed sum is charged as the depreciation

amount throughout the lifetime of an asset such that the accumulated sum at the

end of the life of the asset is exactly equal to the purchase value of the asset.

Here, e make an important assumption that inflation is absent. Let P = first cost of the asset, F = salvage value of the asset n = life of the asset,

Bt = book value of the asset at the end of the period t, Dt = depreciation amount for the period t.

The formulae for depreciation and book value are as follows:

Dt = (P – F)/n

Bt = Bt–1 – Dt = P – t [(P – F)/n]

EXAMPLE

A company has purchased an equipment whose first cost is Rs. 1,00,000 with an

estimated life of eight years. The estimated salvage value of the equip ent at the end of its

lifetime is Rs. 20,000. Determine the depreciation charge and book value at the end of

various years using the straight line method of depreciation.

Solution

P = Rs.

1,00,000 F = Rs.

20,000 n = 8

years Dt = (P – F)/n

= (1,00,000 – 20,000)/8 = Rs. 10,000

In this method of depreci tion, the v lue of Dt is the same for all the years. The

calculations pertaining to Bt for different values of t are summarized in Table .

Table Dt and Bt Va ues under Straight line Method of Depreciation

End of year Depreciation Book value

(t) (D t

) (B t = B t–1

– D t

)

0 1,00,000

1 10,000 90,000

2 10,000 80,000

3 10,000 70,000

4 10,000 60,000

5 10,000 50,000

6 10,000 40,000

7 10,000 30,000

8 10,000 20,000

EXAMPLE

Consider Example and compute the depreciation and the book value for period 5.

P = Rs. 1,00,000

F = Rs. 20,000

n = 8 years

D5 = (P – F)/n

= (1,00,000 – 20,000)/8

= Rs. 10,000 (This is independent f the time period.)

Bt = P – t (P – F)/n

B5 = 1,00,000 – 5 (1,00,000 – 20,000)/8

= Rs. 50,000

2. Declining Balance Method of Depreci tion

In this method of deprecia ion, a constant percentage of the book value of the previous period of the sset will be charged as the

depreciation amount for the current period. This approach is a more realistic approach, since the depreciation

charge decreases with the life of the asset which matches with the earning potent al of the asset.

The book value at the end of the life of the asset may not be exactly equal to the sal age value of the asset. This is a major limitation of this approach.

Let

P = first cost of the asset,

F = salvage value of the asset,

n = life of the asset,

Bt = book value of the asset at the end of the period t,

K = a fixed percentage, and

Dt = depreciation amount at the end of the period t.

The formulae for depreciation and book value are as follows: D

t =

K

B

t-1

Bt =

B

t–1 –

D

t =

B

t–1 –

K

B

t–1

= (1 – K) Bt–1

.The formulae for depreciation and book value in terms of P are as follows:

Dt = K(1 – K)t–1

P Bt = (1 – K)t

P

While availing income-tax exception for the depreciation

amount paid in each year, the rate K is limited to at the ost 2/n. If

this rate is used, then the corresponding approach is called the

double declining balance method of depreciation.

5. Service Output Method of Depreciation

In some situations, it may not be realistic to compute depreciation

based on time period. In such cases, the depreciation is computed

based on service rendered by an asset. Let

P = first cost of the asset

F = salvage value of the asset

X = maximum capacity of service of the asset during its lifeti e

x = quantity of service rendered in a period.

Then, the depreciation is defined per unit of service

rendered: Depreciation/unit of service = (P – F)/X

Depreciation for x units of service in a period = P − F (

x)

X

EXAMPLE

The first coat of a road laying m chine is Rs. 80,00,000. Its salvage value

after five years is Rs. 50,000. The leng h of road that can be laid by the

machine during its lifetime is 75,000 km. In its third year of operation, the

length of road laid is 2,000 km. Find the depreciation of the equipment for

that year.

Solution P = Rs. 80,00,000

F = Rs. 50,000

X = 75,000 km

x = 2,000 km P − F

Depreciation for x units of service in a period = X

(80,00,000 − 50,000)

Depreciation for year 3 =

= Rs. 2,12,000

5.3 Evaluation Of Public Alternatives In evaluating alternatives of private organizations, the criterion is to select the alternative with the maximum profit.

✓ The profit maximization is the main goal of private organizations while providing goods/services as per specifications to their customers.

✓ But the same criterion cannot be used while evaluating public

alternatives. Examples of some public alternatives are

✓ constructing bridges,

✓ roads, dams,

establishing public utilities, etc.

The main objective of any public alternative is to provide goods/services to the public at the minimum cost. In this process, one should see whether the benefits of the

public activity are at least equal to its cos s.

5.4 Inflation Adjusted Decisions

Inflation is the rate of increase in the prices of goods per period. So, it has a compounding effect. Thus, prices that are inflated at a rate of 7% per year will increase

7% in the first year, and for the next year the expected increase will be 7% of these new prices.

If economic decisions are taken without considering the effect of inflation into account, most of them would become meaningless and as a result the organizations

would end up with unpred ctable return.

5.5 Procedure To Adjust Inflation

A procedure to deal with this situation is summarized no .

1. Estimate all the costs/returns associated with an investment proposal in terms of

today’s rupees.

2. Modify the costs/returns estimated in step 1 using an assumed inflation rate so that

at each future date they represent the costs/returns at that date in terms of the rupees that must be expended/received at that time, respectively.

3. As per our requirement, calculate either the annual equivalent amount or future

amount or present amount of the cash flow resulting from step 2 by considering the time value of money.